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4 years ago

What is the area of this composite figure? 96 cm2 160 cm2 208 cm2 256 cm2

Mathematics

2 answers:

gizmo_the_mogwai [7]4 years ago

8 0

The figure is composed by 2 Isosceles Triangles (Two equal sides) and one Rectangle in the middle.

Triangle Area:
$base \times height \div 2$
Rectangle Area:
$length \times width$
- Triangles:
Base: 16cm
Height: 6cm

- Rectangle:
Length: 16cm
Width: 10cm

Triangle Area (At):

- At = 16 x 6/2
- At = 96/2
- At = 48 cm^2

Rectangle Area (Ar):

- Ar = 16 x 10
- Ar = 160 cm^2

Total Area = 2 x (Triangle Area) + Rectangle Area

Total Area = 2 x 48 + 160
Total Area = 96 + 160
Total Area = 256cm^2

Answer: 256cm^2, last alternative.

blondinia [14]4 years ago

3 0

To find the total area of this polygon, you would have to find the areas of the rectangle and two triangles and add them up.

Area of Triangle = 1/2 x base x height
= 1/2 x 16 x 6
= 8 x 6
= 48

And since there are two triangles, we can multiply 48 by 2 to get the areas of both triangles.

Area of Both Triangles = 48 x 2 = 96
Area of Rectangle = base x height
= 16 x 10 = 160

Total Area = 96 + 160 = 256 cm2

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Can someone check my answer?

ivanzaharov [21]

The measure of EC is 1 ft.

Solution:

Given data:

AD = 8 ft, DB = 2 ft, AE = 4 ft

To find the measure of EC:

Using triangle proportionality theorem,

<em>If a line is parallel to one side of a triangle intersects the other two sides, then it divides the two side proportionally.</em>

$$\frac{AD}{DB} =\frac{AE}{EC}$

$$\frac{8}{2} =\frac{4}{EC}$

Do cross multiplication.

$8EC=2\times 4$

8 EC = 8

Divide by 8 on both side of the equation, we get

EC = 1

EC = 1 ft

Hence the measure of EC is 1 ft.

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3 years ago

PLS HELP, RLY URGENT

saul85 [17]

ANSWER:
The right option is D)
A BLACK QUEEN.
Because in first option the cards are red.
In second option the black king can be the answer but a red heart is also there in the option so that is not the answer .
In third there is given no black face card which is not possible as there are only black cards.
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3 years ago

What is the domain and range of each function

yKpoI14uk [10]

Answer:

1. Domain: (3, 5, 7, 8, 11)

Range: ( 6, 7, 9, 14)

2. Domain: (-3, -1, 2, 5, 7)

Range: (-7, -5, -4, 5, 9)

Step-by-step explanation:

The domain is the set of all x values in the function. The range is the set of all y values in the function. It is written in parenthesis from least to greatest. Do not repeat values even if the function has them more than once.

1. Domain: (3, 5, 7, 8, 11)

Range: ( 6, 7, 9, 14)

2. Domain: (-3, -1, 2, 5, 7)

Range: (-7, -5, -4, 5, 9)

6 0

3 years ago

If f(x)=7x+8 and g(x)=x^4, what is (g°f)(0)

Lina20 [59]

G(f(x))= (7x+8)^4
g(f(x))= (7(0)+8)^4
=(0+8)^4
=8^4
=8•8•8•8
=4,096

8 0

3 years ago

Read 2 more answers

A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has fa

gayaneshka [121]

Answer:

the probability the die chosen was green is 0.9

Step-by-step explanation:

Given that:

A bag contains two six-sided dice: one red, one green.

The red die has faces numbered 1, 2, 3, 4, 5, and 6.

The green die has faces numbered 1, 2, 3, 4, 4, and 4.

From above, the probability of obtaining 4 in a single throw of a fair die is:

P (4 | red dice) = $\dfrac{1}{6}$

P (4 | green dice) = $\dfrac{3}{6}$ = $\dfrac{1}{2}$

A die is selected at random and rolled four times.

As the die is selected randomly; the probability of the first die must be equal to the probability of the second die = $\dfrac{1}{2}$

The probability of two 1's and two 4's in the first dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^4$

= $\dfrac{4!}{2!(4-2)!} ( \dfrac{1}{6})^4$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^4$

= $6 \times ( \dfrac{1}{6})^4$

= $(\dfrac{1}{6})^3$

= $\dfrac{1}{216}$

The probability of two 1's and two 4's in the second dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $6 \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $( \dfrac{1}{6}) \times ( \dfrac{3}{6})^2$

= $\dfrac{9}{216}$

∴

The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )

The probability of two 1's and two 4's in both die = $\dfrac{1}{216} \times \dfrac{1}{2} + \dfrac{9}{216} \times \dfrac{1}{2}$

The probability of two 1's and two 4's in both die = $\dfrac{1}{432} + \dfrac{1}{48}$

The probability of two 1's and two 4's in both die = $\dfrac{5}{216}$

By applying Bayes Theorem; the probability that the die was green can be calculated as:

P(second die (green) | two 1's and two 4's ) = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)

P(second die (green) | two 1's and two 4's ) = $\dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}$

P(second die (green) | two 1's and two 4's ) = $\dfrac{0.5 \times 0.04166666667}{0.02314814815}$

P(second die (green) | two 1's and two 4's ) = 0.9

Thus; the probability the die chosen was green is 0.9

8 0

3 years ago